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JEE PHYSICS MCQ: OSCILLATIONS

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Analyzing +4 | -1
Under the influence of a periodic force F1\mathrm{F} 1 a particle performs SHM\mathrm{SHM} along x\mathrm{x}-axis and its x\mathrm{x} coordinate varies with time as, x=Asin⁡ωtx=A \sin \omega t. When the force F1F 1 is replaced by another periodic force F2, particle performs SHM along yy-axis and its yy-coordinate varies with time as, y=Bsin⁡(ωt+Φ)y=B \sin (\omega t+\Phi). When both forces are applied on the particle simultaneously, the particle performs SHM along a straight line in the x−yx-y plane.

(i) The value of Φ\Phi may be zero

(ii) The value of Φ\Phi may be π\pi

(iii) The slope of the resulting path may be BA\frac{B}{A}

(iv) The slope of the resulting path may be −BA-\frac{B}{A}

Choose the correct option.
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Correct option: D
All (i), (ii), (iii) and (iv) are correct.
Option D (All (i), (ii), (iii) and (iv) are correct) is correct because the phase difference can be 0 or π\pi, and the slope of the resulting path can be BA\frac{{B}}{{A}} or −BA-\frac{{B}}{{A}}.

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